Simple $M_n(D)$-module with $D$ a division ring Define $D$ to be a division algebra over a field $k$ and $R=M_n(D)$ the $n\times n$ matrix ring over $D$. A simple $R$-module $M$ is the quotient of $R$. I can write $R=\bigoplus_j I_j$ where $I_j$ is the subring of $R$ all of whose columns except the $j$-th are zero. 
Is it true that the induced map $R\to M$ induces an isomorphism between $M$ and some $I_j$? 
Many thanks!
 A: The answer is that "yes, $M$ is isomorphic to $I_j$ (for any $j$!), but it is not necessarily 'induced' by a projection $R\to M$."
Given any simple left module $S$, a projection $\phi: R\to S$ turns into an isomorphism $R\ker\phi\cong S$. Where $\ker\phi$ is a maximal left ideal of $R$. Since $R$ is semisimple, $R\cong \ker\phi\oplus N$ where $N$ is a left ideal of $R$. So $S\cong R/\ker\phi=(N\oplus\ker\phi)/\ker\phi\cong N$. 
This shows that every simple left module appears as a minimal left ideal of $R$. But $N$ need not be one of the $I_j$. For example, $N$ could be the set of matrices which have an arbitrary first column, the second column identical with the first, and the rest of the columns zero. There are lots of other simple left ideals of $R$ other than the $I_j$.
To prove that all simple left $R$ modules are mutually isomorphic, we can use this
Lemma: For a fixed minimal left ideal $N<R$, the sum $I_N=\sum\{N'<R\mid N\cong N'\}$ is an ideal of $R$ ($N'$ denotes a left ideal, of course.) Furthermore, if $L$ is another minimal left ideal which is nonisomorphic to $N$, then $I_L\cap I_N=\{0\}$.
Since $R$ is simple, only one such $I_N$ can exist, and that means there is only one isotype of simple left $R$ module. That is why all of the $I_j$ in your original post are mutually isomorphic, and indeed every other simple left $R$ module you find is going to be isomorphic to these also.
A: 
$R=M_n(D)$ the $n\times n$ matrix ring over $D$.  $I_j$ is the
subring of $R$ all of whose columns except the $j$-th are zero.

There is another way to prove that any simple left $R$-module is isomorphic to $I_1$. We divide the proof into two parts.

*

*First, we show that any simple left $R$-module is isomorphic to some
simple left $R$-submodule of  $_RR$ (equivalently, minimal left
ideal of $R$).


*Second, we show that any simple left $R$-submodule of $_RR$ is
isomorphic to $I_1$ .
For the first part: let $S$ be a simple left $R$-module. One can check that $S$ is cyclic and say $x$ is its generator. It's easy to construct a non-zero epimorphism $R \to Rx = S$. Since $R$ is semisimple, the epimorphism splits.(See Corollary 13.10 in Rings and Categories of Modules 2ed) Hence $S$ is isomorphic to some simple left $R$-submodule of $_RR$.
For the second part: let $T$ be a simple left $R$-submodule of $_RR$. Since $T\ne0$, there must exist some integer $w$ and matrix $A \in T$ such that the $w$-th colunm of $A$ is not $0$. Without lose of generality, let's say $w = 1$. Then we construct a non-zero $R$-module homorphism $T \to I_1$ that only preserves the first column of matrices in $T$. Since $I_1$ is also simple, by Schur's lemma, we have $T \cong I_1$.
A: This answer does not add anything to rschwieb's one from the theoretical point of view. However, for didactic reasons, I had to prove that any simple $\mathsf{Mat}_n(D)$-module is isomorphic to $D^n$ directly (and by elementary arguments). The following is the argument I found and I think can be interesting for our community herein.
Set $R=\mathsf{Mat}_n(D)$, for the sake of simplicity. Let $V$ be a simple $R$-module. It can be easily checked that there exists $v\neq 0$ in $V$ such that $Rv = V$. In particular, there exists an $E_{ij}$ such that $E_{ij}v \neq 0$, where $E_{ij}$ is the matrix with $1$ in $(i,j)$-th position and $0$ elsewhere. Consider the assignment
$$
\phi:D^n \to V, \qquad \begin{pmatrix}a_1 \\ \vdots \\ a_n\end{pmatrix} \mapsto \sum_k a_kE_{kj}v = 
\underset{j\text{-th column}}{\underbrace{\begin{pmatrix}
0 & \cdots & 0 & a_1 & 0 & \cdots & 0 \\
0 & \cdots & 0 & a_2 & 0 & \cdots & 0 \\
\vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & a_n & 0 & \cdots & 0 \\
\end{pmatrix}}}v.
$$
We have that for every $\begin{pmatrix}a_1 \\ \vdots \\ a_n\end{pmatrix} \in D^n$,
$$
\phi\left((x_{ij})\begin{pmatrix}a_1 \\ \vdots \\ a_n\end{pmatrix}\right) = \phi\begin{pmatrix} \sum_i x_{1,i}a_i \\ \vdots \\ \sum_i x_{n,i}a_i\end{pmatrix} = \sum_k \sum_i x_{k,i}a_iE_{kj}v = \left((x_{ij})\left(\sum_k a_kE_{kj}\right)\right)v = (x_{ij})\left(\left(\sum_k a_kE_{kj}\right)v\right) = (x_{ij})\phi\begin{pmatrix}a_1 \\ \vdots \\ a_n\end{pmatrix},
$$
that is to say, $\phi$ is $R$-linear. In addition, $\phi$ is not zero (because $\phi(e_i) = E_{ij}v \neq 0$) and it is an $R$-module homomorphism between simple $R$-modules, whence it has to be an isomorphism.
